Question

find the range of k if the quadratic equation
2x^2-x=k has real and distinct roots.

Answer 1

-b/2a will give you the lowest point

Answer 2

or rather, when x= -b/2a

thats the one lol

Answer 3

so the answer is?

Answer 4

that is the answer .....

Answer 5

oh god, just do b^2 - 4ac.

(-1)^2 - (4 * 2 * -k) < 0

so 1 + 8k < 0

so k< (-1/8)

Answer 6

lol okay thanks :)

Answer 7

@tanvidais, shoudn't it be b^2-4ac = 0 ?

Answer 8

because it has real and distinct roots right?

Answer 9

um if the discriminant is <0 it has no real roots ,=0 has 1 real root, >0 has 2 real roots i thought

Answer 10

and it says it has real and distinct roots. so the discriminant supposed to be =0 right?

Answer 11

the range of: 2x^2-x = k is no different from the range of: 2x^2-x=y

2x^2-x = k

2(x^2-1/2 x) = k

2(x^2-1/2 x + 1/16) - 1/8 = k

2(x-1/4)^2 - 1/8 = k

it looks to be then from -1/8 to infinity if im reading it right

Answer 12

oh, shoot. then it's k> (-1/8)

Answer 13

>0 since it has 2 diff roots it says in question "real and distinct"

Answer 14

when it is real roots, it is => 0.

when it is real and distinct, it is > 0.

when it is equal roots, it is = 0.

when it is no roots, it is <0.

Answer 15

thanks! :)

Answer 16

no problem :)

Answer 17

2x^2 -x -k ; such that k > -1/8 has 2 real and disticnt roots